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Re: Every set x equinumerous with a set y disjoint from x?

Subject: Re: Every set x equinumerous with a set y disjoint from x?
From: "Rupert"
Date: 28 Aug 2006 17:11:56 -0700
Newsgroups: sci.logic
MoeBlee wrote:
> Rupert wrote:
> > That looks good to me. David's proof looks good to me as well.
>
> So would I be correct to take it that the consensus is that the theorem
> almost surely needs choice?
>

No, David's proof is correct and doesn't use choice.

> If that is the case, I find it interesting that this is not mentioned
> much (if at all) in basic textbooks. It seems so intuitive that for any
> set there is an equinumerous set disjoint from the original set, so the
> need for the axiom of choice would be another in the list of "reasons"
> for the axiom.
> 
> MoeBlee


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